Question: Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}5x-8y &= 4 \\ -5x+6y &= -8\end{align*}$
Answer: Begin by moving the $x$ -term in the second equation to the right side of the equation. $6y = 5x-8$ Divide both sides by $6$ to isolate $y$ $y = {\dfrac{5}{6}x - \dfrac{4}{3}}$ Substitute this expression for $y$ in the first equation. $5x-8({\dfrac{5}{6}x - \dfrac{4}{3}}) = 4$ $5x - \dfrac{20}{3}x + \dfrac{32}{3} = 4$ Simplify by combining terms, then solve for $x$ $-\dfrac{5}{3}x + \dfrac{32}{3} = 4$ $-\dfrac{5}{3}x = -\dfrac{20}{3}$ $x = 4$ Substitute $4$ for $x$ back into the top equation. $5( 4)-8y = 4$ $20-8y = 4$ $-8y = -16$ $y = 2$ The solution is $\enspace x = 4, \enspace y = 2$.